--- title: "Industry room 30 × 20 m LED lighting and DB" subject: "Industrial Power Distribution and Illumination" chapter: 7 key: "semester_6/industrial_power_distribution_and_illumination:industry-room-30x20-led" source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/industry-room-30x20-led site: BCT Engineering --- # Industry room 30 × 20 m LED lighting and DB Industrial Power Distribution and Illumination · Chapter 7: Illumination Design Principles - Asked in: 2074 Bhadra, 12 marks - Video: [How to solve illumination problem's !! question from VTU QP !! By Sharan Sir!!](https://www.youtube.com/watch?v=-pIazs-fjYs) — Knowledge center - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/industry-room-30x20-led - Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-7 ## Question The size of the industry room is 30m × 20m and it is illuminated by 42 W LED lamp. The efficiency of lamp is 110 lumen/watt and the distance between working plane and mounting height of luminaries is 3 meters. The coefficient of utilization and depreciation factor are 0.75 and 1.2 respectively. Calculate the number of luminaries for the industrial room. Draw neat sketch showing the arrangement of lamps, switches, power sockets and decide the light and power sub-circuits. Design the distribution board to supply lighting loads and power if the supply system is 3 phase, 400V and 50 Hz. ## Short answer 63 LED luminaires (9×7 grid) provide 303 lux. Nine 6A lighting sub-circuits (7 luminaires each) and six 20A power sub-circuits (2 sockets each) balanced across three phases. DB: 32A TPN MCB incomer, 40A 4P RCCB (100mA), 6A/20A SP MCBs, 4-core 6mm² Cu incoming cable. ## Answer **Given:** room 30 m × 20 m, 42 W LED at 110 lm/W, Hm = 3 m, CU = 0.75, depreciation factor 1.2, supply 400 V, 3-phase, 50 Hz. **Assumptions:** the required illumination is not given, so **E = 300 lux** is taken (medium industrial work, IS 3646); SHR ≤ 1.5; pf 0.9; 12 power sockets of 16 A (1 kW each). #### Number of luminaires Lumen method: N = (E × A) / (φ × UF × MF), where φ is the output of one luminaire. ```text Area A = 30 × 20 = 600 m² φ per luminaire = 42 W × 110 lm/W = 4620 lm MF = 1/DF = 1/1.2 = 0.833 Total lumens = E × A × DF / UF = 300 × 600 × 1.2 / 0.75 = 288000 lm N = 288000 / 4620 = 62.34 ``` Rounding up and fitting a regular grid: **9 × 7 = 63 luminaires** (9 along the 30 m side, 7 along the 20 m side). #### Arrangement and spacing ```text Spacing along length = 30/9 = 3.33 m Spacing along width = 20/7 = 2.86 m End spacing (to wall) = half of these Max spacing = SHR × Hm = 1.5 × 3 = 4.5 m 3.33 m and 2.86 m ≤ 4.5 m → OK Actual E = 63 × 4620 × 0.75 × 0.833 / 600 = 303.2 lux (≥ 300 lux required) ``` #### Layout diagram ```text Plan 30 m × 20 m (letter = phase of lamp circuit) +-------------------+ | R R R R R R R Y Y | | B B B B Y Y Y Y Y | | B B B R R R R R R | | B Y Y Y Y Y Y Y R | | B B B B B B R R R | | Y Y Y Y Y R R R R | | Y Y B B B B B B B | +-------------------+ [DB] [SB] <- entrance: DB and switch board S S S S ... one switch per sub-circuit ``` Each light sub-circuit is controlled by one 6 A one-way switch; all switches are grouped on a switch board (SB) next to the entrance, and the DB is mounted beside it. Power sockets (16 A, 3-pin, with earth) are placed along the walls at about 1 m height, spaced roughly evenly. #### Sub-circuits Rule (IS 732 / NEA practice): a light sub-circuit carries at most 10 points and 800 W. Here one luminaire is 42 W, so up to 10 luminaires per sub-circuit. Number of circuits is made a multiple of 3 for phase balance. | Circuit | Luminaires | Load (W) | Current (A) | Phase | MCB | |---|---|---|---|---|---| | L1 | 7 | 294 | 1.42 | R | 6 A | | L2 | 7 | 294 | 1.42 | Y | 6 A | | L3 | 7 | 294 | 1.42 | B | 6 A | | L4 | 7 | 294 | 1.42 | R | 6 A | | L5 | 7 | 294 | 1.42 | Y | 6 A | | L6 | 7 | 294 | 1.42 | B | 6 A | | L7 | 7 | 294 | 1.42 | R | 6 A | | L8 | 7 | 294 | 1.42 | Y | 6 A | | L9 | 7 | 294 | 1.42 | B | 6 A | Power sub-circuits (assumed): 12 sockets of 16 A, each taken as 1 kW, at most 2 sockets per power sub-circuit (IS 732), 2.5 mm² Cu wire, 20 A MCB. | Circuit | Sockets | Load (W) | Phase | MCB | |---|---|---|---|---| | P1 | 2 | 2000 | R | 20 A | | P2 | 2 | 2000 | Y | 20 A | | P3 | 2 | 2000 | B | 20 A | | P4 | 2 | 2000 | R | 20 A | | P5 | 2 | 2000 | Y | 20 A | | P6 | 2 | 2000 | B | 20 A | #### Distribution board design | Phase | Circuits | Load (W) | Current (A) | |---|---|---|---| | R | L1, L4, L7, P1, P4 | 4882 | 23.58 | | Y | L2, L5, L8, P2, P5 | 4882 | 23.58 | | B | L3, L6, L9, P3, P6 | 4882 | 23.58 | ```text Connected load = 2646 W light + 12000 W power = 14646 W Most loaded phase = 4882 W Iph = 4882 / (230 × 0.9) = 23.58 A Main MCB ≥ 1.25 × 23.58 = 29.48 A → 32 A TPN MCB ``` **DB specification:** 6-way-per-phase TPN DB (18 ways in all, spares included) with 32 A TPN MCB as incomer, 40 A 4-pole RCCB (100 mA) for earth-leakage protection, 6 A SP MCBs (B-curve) for light circuits and 20 A SP MCBs for power circuits. Incoming cable: 4-core 6 mm² Cu PVC (rating 45 A ≥ 32 A), plus earth continuity conductor. ```text 3-phase, 400/230 V, 4-wire supply | [Energy meter] | [32 A TPN MCB] | [40 A 4P RCCB] | ==R====Y====B====N== busbars | | | | | L1 L2 L3 P1 P2 ... ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the output of one 42 W, 110 lm/W LED luminaire found? A. 42 / 110 = 0.38 lm B. 42 x 110 = 4620 lm C. 42 + 110 = 152 lm D. 110 / 42 = 2.6 lm Answer: B. Lumen output = power x luminous efficacy = 42 x 110 = 4620 lm. (Key point: Luminous flux per lamp) 2. The depreciation factor is given as 1.2. How does it enter the lumen method? A. MF = 1.2 is used directly in the denominator B. MF = 1/DF = 0.833, so total lumens = E x A x DF / UF C. It is subtracted from UF D. It is ignored because it is above 1 Answer: B. Maintenance factor is the reciprocal of the depreciation factor, so DF multiplies the required lumens. (Key point: Depreciation factor and MF) 3. The question gives no required lux. What does the answer do? A. Takes E = 300 lux for medium industrial work (IS 3646) and states it as an assumption B. Refuses to calculate C. Takes E = 100 lux for all rooms D. Takes E from the lamp wattage Answer: A. A missing data item is assumed from standard practice and stated clearly; here 300 lux. (Key point: Assumed illuminance) 4. Total lumens = 300 x 600 x 1.2 / 0.75 = 288000. How many luminaires are fitted? A. 62 in a 31 x 2 grid B. 63 in a 9 x 7 grid, giving about 303 lux C. 70 in a 10 x 7 grid D. 288000 / 42 = 6857 Answer: B. 288000 / 4620 = 62.34, rounded up to 63 = 9 x 7, with actual E = 303.2 lux. (Key point: Number of luminaires) 5. How are the 12 power sockets of 16 A arranged? A. All 12 on one 20 A circuit B. At most 2 sockets per circuit, giving 6 circuits of 20 A, two per phase C. One socket per light circuit D. Each socket on a 6 A MCB Answer: B. IS 732 limits sockets per power sub-circuit to 2, so 6 circuits (P1-P6) are spread evenly over the phases. 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